Comments on Why $\color{red}{k\dbinom{k}{1}} \neq$ "first choose the k team members and then choose one of time to be captain"?
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Why $\color{red}{k\dbinom{k}{1}} \neq$ "first choose the k team members and then choose one of time to be captain"?
Because you "first choose the k team members and then choose one of time to be captain", shouldn’t the RHS be $\color{red}{k\dbinom{k}{1}}$? The captain is chosen from the $k$ team members already chosen.
$\color{forestgreen}{k\dbinom{n}{k}}$ appears wrong to me, because this means that you're choosing the captain from the original group of $n$ people.
Example 1.5.2 (The team captain).
For any positive integers n and k with $k \le n$, $n\dbinom{n - 1}{k - 1} = \color{forestgreen}{k\dbinom{n}{k}}$.
This is again easy to check algebraically (using the fact that $m! = m(m - 1)!$ for any positive integer $m$), but a story proof is more insightful.
Story proof : Consider a group of n people, from which a team of k will be chosen, one of whom will be the team captain. To specify a possibility, we could first choose the team captain and then choose the remaining $k - 1$ team members; this gives the left-hand side. Equivalently, we could first choose the k team members and then choose one of them to be captain; this gives the right-hand side. $\square$
Blitzstein. Introduction to Probability (2019 2 ed). Pages 20-21.
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$(^n_k)$ is the number of ways to choose $k$ team members from $n$ players.
$k$ is the number of ways to choose one captain from $k$ team members. (It isn't the number of ways to choose a captain from the original group of people, as you suggested; that would be $n$.)
$k(^n_k)$ is the product of the two, which is the number of ways to do both independently.
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